\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\) [165]

Optimal result

Integrand size = 33, antiderivative size = 254 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 (17 A-7 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

7/10*(17*A-7*B)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(33*A-13*B 
)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d+1/6*(33*A-13*B)*sin(d*x+c)/ 
a^3/d/cos(d*x+c)^(3/2)-7/10*(17*A-7*B)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)-1 
/5*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3-1/3*(2*A-B)*sin( 
d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-7/30*(17*A-7*B)*sin(d*x+c)/ 
d/cos(d*x+c)^(3/2)/(a^3+a^3*cos(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 9.31 (sec) , antiderivative size = 1110, normalized size of antiderivative = 4.37 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3) 
,x]
 

Output:

(-22*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, 
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S 
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa 
n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^ 
3*Sqrt[1 + Cot[c]^2]) + (26*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hypergeometric 
PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc 
Tan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2] 
*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/( 
3*d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqr 
t[Cos[c + d*x]]*((-2*(60*A - 20*B + 59*A*Cos[c] - 29*B*Cos[c])*Csc[c/2]*Se 
c[c/2]*Sec[c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(59*A*Sin[(d*x)/2] - 
 29*B*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(16*A*Sin[(d 
*x)/2] - 11*B*Sin[(d*x)/2]))/(15*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A* 
Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(5*d) + (16*A*Sec[c]*Sec[c + d*x]^2*Sin[d* 
x])/(3*d) + (16*Sec[c]*Sec[c + d*x]*(A*Sin[c] - 9*A*Sin[d*x] + 3*B*Sin[d*x 
]))/(3*d) - (4*(16*A - 11*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A 
 - B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3 - (119 
*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4} 
, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(S 
qrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*...
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3457, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B)-25 a^2 (2 A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B)-25 a^2 (2 A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \cos (c+d x)\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-7 a^3 (17 A-7 B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-7 a^3 (17 A-7 B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-7 a^3 (17 A-7 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-7 a^3 (17 A-7 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-7 a^3 (17 A-7 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-7 a^3 (17 A-7 B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\)

Input:

Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3),x]
 

Output:

-1/5*((A - B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3) 
+ ((-10*a*(2*A - B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d 
*x])^2) + ((-7*a^2*(17*A - 7*B)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a 
*Cos[c + d*x])) + (3*(5*a^3*(33*A - 13*B)*((2*EllipticF[(c + d*x)/2, 2])/( 
3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) - 7*a^3*(17*A - 7*B)*((- 
2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))) 
)/(2*a^2))/(3*a^2))/(10*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(875\) vs. \(2(233)=466\).

Time = 6.78 (sec) , antiderivative size = 876, normalized size of antiderivative = 3.45

method	result	size
default	\(\text {Expression too large to display}\)	\(876\)

Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

1/60*(4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2* 
d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(165*A*EllipticF(cos(1/2* 
d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 
)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-10*(2*sin(1/2*d*x+1/2*c)^2-1)^ 
(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE 
(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+14 
7*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos(1/2*d* 
x+1/2*c)+8*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(165*A*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(2*sin(1/2*d*x+1/2*c)^2-1 
)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 
147*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-168*(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(17*A-7*B)*sin(1/2*d*x+1/2* 
c)^10+8*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(1167*A-48...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 548, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

-1/60*(2*(21*(17*A - 7*B)*cos(d*x + c)^4 + 2*(453*A - 188*B)*cos(d*x + c)^ 
3 + 5*(139*A - 59*B)*cos(d*x + c)^2 + 60*(2*A - B)*cos(d*x + c) - 20*A)*sq 
rt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^ 
5 + 3*sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(33*I*A - 13*I* 
B)*cos(d*x + c)^3 + sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^2)*weierstrassP 
Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-33*I*A + 13*I 
*B)*cos(d*x + c)^5 + 3*sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c)^4 + 3*sqrt( 
2)*(-33*I*A + 13*I*B)*cos(d*x + c)^3 + sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x 
+ c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(sq 
rt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^5 + 3*sqrt(2)*(-17*I*A + 7*I*B)*cos(d 
*x + c)^4 + 3*sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(-17*I*A 
+ 7*I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + 
c)^5 + 3*sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(17*I*A - 7*I 
*B)*cos(d*x + c)^3 + sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^2)*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^ 
3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3 
*d*cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorith 
m="giac")
 

Output:

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^3*cos(d*x + c)^(5/2)) 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^3),x)
                                                                                    
                                                                                    
 

Output:

int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^3), x)
 

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}+3 \cos \left (d x +c \right )^{5}+3 \cos \left (d x +c \right )^{4}+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5}+3 \cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+\cos \left (d x +c \right )^{2}}d x \right ) b}{a^{3}} \] Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x)
 

Output:

(int(sqrt(cos(c + d*x))/(cos(c + d*x)**6 + 3*cos(c + d*x)**5 + 3*cos(c + d 
*x)**4 + cos(c + d*x)**3),x)*a + int(sqrt(cos(c + d*x))/(cos(c + d*x)**5 + 
 3*cos(c + d*x)**4 + 3*cos(c + d*x)**3 + cos(c + d*x)**2),x)*b)/a**3